(a) The diagram shows the graphs of
y = 2x and
y = 2x + 1 .
(i) State the gradient of the line
y = 2x + 1 .
(ii) Find the value of x such that
x > 0 and 2x + 1 = 2x .
Answer (a)(i) .............................................. [1]
(ii) x = ....................................... [1]
(b) The diagram shows the graph of
y = kax .
State the value of
(i) k ,
(ii) a.
Answer (b)(i) k = ....................................... [1]
(ii) a = ....................................... [1]
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For
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Let's break down the problem step by step.

### Part (a)
We are given two equations:
1. \( y = 2x \)
2. \( y = 2x + 1 \)

#### (a)(i) Gradient of the line \( y = 2x + 1 \)
The general form of a linear equation is \( y = mx + c \), where:
- \( m \) is the gradient (slope),
- \( c \) is the y-intercept.

For the equation \( y = 2x + 1 \), we can directly see that the gradient \( m = 2 \).

**Answer (a)(i): The gradient is 2.**

#### (a)(ii) Solve \( 2x + 1 = 2x \)
We need to find the value of \( x \) such that \( x > 0 \) and \( 2x + 1 = 2x \).

1. Subtract \( 2x \) from both sides:
   \[
   2x + 1 - 2x = 2x - 2x
   \]
   This simplifies to:
   \[
   1 = 0
   \]

This is a contradiction, which means there is **no value of \( x \)** that satisfies the equation \( 2x + 1 = 2x \) when \( x > 0 \).

**Answer (a)(ii): There is no solution for \( x > 0 \).**

### Part (b)
We are given the equation \( y = k a^x \), and we need to find the values of \( k \) and \( a \).

#### (b)(i) Value of \( k \)
\( k \) represents the initial value when \( x = 0 \). Based on the diagram (which I assume shows an exponential graph), the value of \( y \) when \( x = 0 \) is the constant \( k \). From the general form \( y = k a^x \), substitute \( x = 0 \):
\[
y = k a^0 = k(1) = k
\]
So, \( y(0) = k \), which is the y-intercept of the graph. 

You would typically find this value from the graph (not given in the problem), but it’s the initial value when \( x = 0 \).

**Answer (b)(i): The value of \( k \) depends on the y-intercept, which can be read off from the graph.**

#### (b)(ii) Value of \( a \)
\( a \) is the base of the exponential function, which can be determined by observing how the graph increases. If the graph's shape shows rapid growth, \( a > 1 \), and if it shows decay, \( 0 < a < 1 \). The exact value of \( a \) also needs to be read off the graph based on the nature of the curve.

**Answer (b)(ii): The value of \( a \) depends on the growth rate of the exponential function, which is read from the graph.**

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Would you like me to go into more details on any part of this explanation or help with another example?

### 5 related questions:
1. What is the difference between a gradient and y-intercept in linear equations?
2. How do we identify contradictions in algebraic equations?
3. How does the value of \( a \) affect the shape of an exponential graph?
4. What happens to an exponential graph when \( k \) is negative?
5. Can the equation \( 2x + 1 = 2x \) have a solution if we remove the constraint \( x > 0 \)?

### Tip:
When solving linear equations, always check for contradictions or undefined results when simplifying!

(a) The diagram shows the graphs of y = 2x and y = 2x + 1.
(i) State the gradient of the line y = 2x + 1.
(ii) Find the value of x such that x > 0 and 2x + 1 = 2x.
Answer (a)(i) .............................................. [1]
(ii) x = ....................................... [1]
(b) The diagram shows the graph of y = kax.
State the value of (i) k, (ii) a.
Answer (b)(i) k = ....................................... [1]
(ii) a = ....................................... [1]

Explore how to solve the linear equation y = 2x + 1 by determining its gradient and solving for x, as well as analyzing the exponential function y = kax. Learn key concepts such as the slope-intercept form and properties of exponential growth. Suitable for Grades 9-10.

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